Electric potential at the center of a insulating sphere

AI Thread Summary
The discussion revolves around calculating the electric potential at the center of a uniformly charged insulating sphere with a given radius and charge. The electric field at r=R/2 is derived using the formula E=(K)(Q)/(2)(R^2). For the potential at r=0, participants clarify that it can be computed by considering the potential on the surface of the sphere, which is given by V=(K)(Q)/R. Confusion arises regarding the integration limits leading to a negative sign in the calculations, but it is resolved by correctly setting up the integrals. Ultimately, the correct potential is confirmed to be positive, aligning with the expected theoretical results.
SpringWater
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Homework Statement



Consider a uniformly charged insulating sphere with radius R and total charge Q in- side the sphere.

If Q = 2.9 × 10−6 C, what is the magnitude of the electric field at r=R/2 K=columbs constant. The answer to this question is E=(K)(Q) / (2)*(R)^(2)

The second part of the question which I am having trouble with is...

If the sphere has a radius of 3.3 m, find the potential at r = 0. The Coulomb constant is 8.98764 × 109 N · m2/C2 . Follow the convention that the electric potential at r = ∞ is zero. answer in units of V

Homework Equations


Q=2.9E-6 C
r=3.3m
k=8.98764E9 N*m^2 / C^2


The Attempt at a Solution



So I have set up the problem as two integrals

that are shown in the picture. I thought that this would give me the electric potential at r=0 however i am still getting it incorrect. I am not sure what I am doing wrong any help would be appreciated.
Thank you,
 

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Have you tried using the fact that the potential inside a spherical shell is constant and is equal to the potential on its surface?
 
Sunil Simha said:
Have you tried using the fact that the potential inside a spherical shell is constant and is equal to the potential on its surface?


I understood that concept but i guess because this was a two part problem i was trying to incorporate an integral that would relate to the first part of the problem, so no i really didnt...

..in that case the electric potential on the surface would be = (K)*(Q) / (R) Correct?

where R is the radius of the sphere and R=3.3 m
 
SpringWater said:
I understood that concept but i guess because this was a two part problem i was trying to incorporate an integral that would relate to the first part of the problem, so no i really didnt...

..in that case the electric potential on the surface would be = (K)*(Q) / (R) Correct?

where R is the radius of the sphere and R=3.3 m

no this is incorrect, so looking back at my integral, i am more sure that my set up was correct, in the orignal post w/ picture. the question now i believe is the sign (-) or (+)?
 
The second part can be solved by assuming the sphere to be made of thin concentric spherical shell of uniform charge densities. Every shell contributes a potential equal to that on it's surface. The solution is simply obtained by integrating the potentials. Note that Q is evenly distributed throughout the sphere and hence it helps to assign a constant as charge density and then use it.
 
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Sunil Simha said:
The second part can be solved by assuming the sphere to be made of thin concentric spherical shell of uniform charge densities. Every shell contributes a potential equal to that on it's surface. The solution is simply obtained by integrating the potentials.

okay, thank you for the reply i greatly appreciate it.

using the integrals i set up, after simplifying the two i obtain -((3)*(k)*(Q) / (2)*(R)) which is a similar formula i just found a min ago however there formula did not have a negative sign. so why when my method is correct, do i come up with a negative sign assuming that the integral sums up the shells.
 
I'm a bit confused about how you got your limits of integration. Would you mind explaining? What is the final answer to this problem, by the way?
 
Averki said:
I'm a bit confused about how you got your limits of integration. Would you mind explaining? What is the final answer to this problem, by the way?


well i set them up incorrectly that is why i kept coming up with a negative answer. see picture for how i switched them. the worked integral is in the first attached picture however the answer should be positive.
 
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Is the answer 3Q/8πεR? If correct I'll explain my method, else I'll check what I've done wrong and later suggest a solution.
 
  • #10
SpringWater said:
well i set them up incorrectly that is why i kept coming up with a negative answer. see picture for how i switched them. the worked integral is in the first attached picture however the answer should be positive.

sorry i forgot the picture.
 

Attachments

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  • #11
Sunil Simha said:
Is the answer \frac{3Q}{8πεR}. If correct I'll explain my method, else I'll check what I've done wrong and later suggest a solution.

yes, i get the same answer from your (simplified) solution and the same from the two integrals i set up however my problem was i did not set up the limits of integration correctly hence the negative sign.

thank you for your help
 
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